Question

Given that, x'2+2x-3 is a factor of f(x)=x'4+6x'3+2ax'2+bx-3a. Find the values of a and b.

Answer 1

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GIVEN,

$\text{x^2+2x+3 is a factor of f(x)=x^4+6x^3+2ax^2+bx-3a}$

First Factorise the divided.

$x^2+3x-x-3$

$x(x+3) -1(x+3)$

$(x-1) (x+3)$

Now,

x+1=0

x=-1

$\text{Put the value of x in f(x)}$

$\implies{ (1)^4+6(1)^3+2a(1)^2+b(1)-3a}$

$\implies{1+6+2a+b-3a}$

$\implies{7-a+b=0}$..........1

Now,

x+2=0

x=-2

$\implies{ (-3)^4+6(-3)^3+2a(-3)^2+b(-3)-3a}$

$\implies{81-162+18a-3b-3a}$

$\implies{ 15a-3b-81}$...........2

Now,

$\text{Multiplying eq 1 with 3}$

We have ,

$3b-3a+21$

add this eqn1 in eqn2

$12a-60=0$

$12a=60$

$a=5$

put value of a in eqn1...we have

$7-5+b=0$

$b=-2$

Thank You!